Answer:
<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>
Explanation:
This question is about solubility.
Regarding solubility, the solutions may be classified as:
- Unsaturated: the concentration is below the maximum concentration permited at the given temperature.
- Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.
- Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.
Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.
- In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.
- In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g of water.
- Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ
115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O ⇒ x = 57.5 g NaNO₃
- <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>
<em />
Most atoms have three different subatomic particles inside them: protons, neutrons, and electrons. The protons and neutrons are packed together into the center of the atom(which is called the nucleus) and the electrons, which are very much smaller, whizz around the outside. Most of an atom is empty space.
I hope this helps you! :D
Answer:
0.55 mol Au₂S₃
Explanation:
Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.
1. Gather all the information in one place:
M_r: 34.08
Au₂S₃ + … ⟶ 3H₂S + …
m/g: 56
2. Calculate the moles of H₂S
Moles of H₂S = 56 g H₂S × (34.08 g H₂S/1 mol H₂S)
= 1.64 mol H₂S
3. Calculate the moles of Au₂S₃
The molar ratio is 1 mol Au₂S₃/3 mol H₂S.
Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)
= 0.55 mol Au₂S₃
So its temperature will not rise, since kinetic energy of molecules remains the same. The quantity of heat absorbed or released when a substance changes its physical phase at constant temperature (e g. From solid to liquid at melting point or from liquid to gas at boiling point) is termed as its latent heat.